IAI Actuarial Core Principles · Actuarial Statistics · Jointly distributed random variables
X and Y have joint density f(x,y) = 2 for 0 < x < y < 1, and 0 otherwise. What is E[X | Y = 0.6]?
Given Y = 0.6, X is uniformly distributed on (0, 0.6) because the joint density is constant over the triangle. Its mean is half the upper limit, which is 0.30.
- A0.20
- B0.60
- C0.25
- D0.40
- 0.30Correct
Explanation
Given Y = y, the density of X is f(x,y)/f_Y(y) = 2/(2y) = 1/y on (0,y), so X is uniform on (0,y). Hence E[X|Y=y] = y/2 = 0.30. The option 0.60 uses y rather than y/2.
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